Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $k \neq 0$. $t = \dfrac{-6}{10(3k + 2)} \times \dfrac{18k^2 + 12k}{8k} $
Solution: When multiplying fractions, we multiply the numerators and the denominators. $t = \dfrac{ -6 \times (18k^2 + 12k) } { 10(3k + 2) \times 8k } $ $ t = \dfrac {-6 \times 6k(3k + 2)} {8k \times 10(3k + 2)} $ $ t = \dfrac{-36k(3k + 2)}{80k(3k + 2)} $ We can cancel the $3k + 2$ so long as $3k + 2 \neq 0$ Therefore $k \neq -\dfrac{2}{3}$ $t = \dfrac{-36k \cancel{(3k + 2})}{80k \cancel{(3k + 2)}} = -\dfrac{36k}{80k} = -\dfrac{9}{20} $